Self-Test 6

1. Two boys take turns throwing darts at a target. Al throws first and hits with probability ¼. Bob throws second and hits with probability 1/3. What is the probability that Al hits the target before Bob does?Solution:¼ + (3/4)(2/3)(1/4) +(3/4)(2/3)(3/4)(2/3)(1/4)+…=(1/4)Sum[(1/2)n,{n,0,}]=1/2.

2. Three girls take turns shooting a basketball and have probabilities 0.2, 0.3 and 0.5 of scoring. Who has the best chance of getting the first basket?

Solution:

(girl 1) 0.2 + (0.8)(0.7)(0.5)0.2+ (0.8*0.7*0.5)20.2 +…=0.278;

(girl 2) 0.8*0.3 + 0.8*0.7*0.5*0.8*0.3 + … 0.8*(0.7*0.5*0.8)2*0.3 +…= 0.24*Sum[(0.7*0.5*0.8)n,{n,0,}]=0.333;

(girl 3) 0.56*0.5 Sum[(0.7*0.5*0.8)n,{n,0,}]=0.389

Check: 0.278+0.333+0.389 =1.

3. A young lady wants to make her date wait occasionally but does not want him to get too angry. Therefore, if she made him wait the previous time, she is only late with probability 0.2, but if she was not late the previous time, she is late with probability 0.6. Let pk be the probability that she is late on the kth date. And suppose that p1 = 1. Compute p1,p2,

and pk.

Solution:

It’s a typical Markov chain.

p[k]= 0.6 (1-p[k-1])+0.2 p[k];

r=0.429

4. Suppose that a rainy day is followed by another rainy day with probability ½, but a sunny day is followed by a sunny day with probability 2/3. What is the long-run fraction of sunny days in this simple-minded model of the weather?

Solution:

p[k] =(2/3)*p[k-1] +0.5*(1-p[k-1]). r=0.5/(1/3+0.5)=0.6

5. 5% of men and 0.25% of women are color-blind. What is a probability a color-blind person is a man?

Solution:

A= “person is color-blind”

B= “a man is color-blind”

P(A)= ½*0.05 + ½*0.0025= 0.02625

P(B|A) = 0.025/( 0.025+0.00125) = 0.952

6. Three factories manufacture 20%, 30% and 50% of the computer chips a company sells. If the fractions of defective chips are 0.4%, 0.3% and 0.2% respectively, what fraction of the defective chips come from the third factory?

Solution:

B= “a chip is defective”; A3 = “a defective chip comes from the third factory”

P(A3|B)= P(A3*B)/P(B)=0.5*0.002/(0.2*0.004+0.3*0.003+0.5*0.002)=0.37

7. Three bags lie on the table. One has two gold coins, one has two silver coins, and one has one silver and one gold.You pick a bag at random and pick out one coin. If this coin is gold, what is the probability you picked from the bag with two gold coins.

Solution:1/3/(1/3+1/3*0.5)=2/3.